Key Concepts & Congruence Rules

Before jumping into the solutions, let us review the essential rules used in this exercise:

• SSS Congruence Rule: If three sides of one triangle are equal to the corresponding three sides of another triangle, then the two triangles are congruent.
• RHS Congruence Rule: If in two right-angled triangles, the hypotenuse and one side of one triangle are equal to the hypotenuse and the corresponding side of the other triangle, then the two triangles are congruent.
• Median of a Triangle: A line segment joining a vertex to the midpoint of the opposite side. It divides the opposite side into two equal halves.
• Altitude of a Triangle: A perpendicular line segment drawn from a vertex to the opposite side.

Question 1

ΔABC and ΔDBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC. If AD is extended to intersect BC at P, show that:
(i) ΔABD ≅ ΔACD
(ii) ΔABP ≅ ΔACP
(iii) AP bisects ∠A as well as ∠D
(iv) AP is the perpendicular bisector of BC.

Solution:

Given: ΔABC is isosceles with AB = AC. ΔDBC is isosceles with DB = DC.

(i) To prove: ΔABD ≅ ΔACD

In ΔABD and ΔACD:

• AB = AC (Given)
• BD = CD (Given)
• AD = AD (Common side)

Therefore, by SSS congruence rule, ΔABD ≅ ΔACD. (Hence Proved)

(ii) To prove: ΔABP ≅ ΔACP

From part (i), since ΔABD ≅ ΔACD, we have:
∠BAD = ∠CAD (by CPCT)
This can also be written as ∠BAP = ∠CAP.

Now, in ΔABP and ΔACP:

• AB = AC (Given)
• ∠BAP = ∠CAP (Proved above)
• AP = AP (Common side)

Therefore, by SAS congruence rule, ΔABP ≅ ΔACP. (Hence Proved)

(iii) To prove: AP bisects ∠A as well as ∠D

Since ∠BAP = ∠CAP, AP bisects ∠A.

Now, let us prove AP bisects ∠D. In ΔBDP and ΔCDP:

• BD = CD (Given)
• DP = DP (Common side)
• BP = CP (Since ΔABP ≅ ΔACP, by CPCT)

By SSS congruence rule, ΔBDP ≅ ΔCDP.
Thus, ∠BDP = ∠CDP (by CPCT). Hence, AP bisects ∠D.

(iv) To prove: AP is the perpendicular bisector of BC

Since ΔABP ≅ ΔACP, we have:

• BP = CP (by CPCT)
• ∠APB = ∠APC (by CPCT)

Also, ∠APB + ∠APC = 180° (Linear pair of angles)
⇒ 2 ∠APB = 180° ⇒ ∠APB = 90°.

Since BP = CP and ∠APB = 90°, AP is the perpendicular bisector of BC. (Hence Proved)

Question 2

AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that:
(i) AD bisects BC
(ii) AD bisects ∠A.

Solution:

Given: ABC is an isosceles triangle with AB = AC, and AD ⊥ BC (∠ADB = ∠ADC = 90°).

In right-angled triangles ΔADB and ΔADC:

• ∠ADB = ∠ADC = 90° (Given AD is an altitude)
• AB = AC (Hypotenuse, Given)
• AD = AD (Common Side)

By RHS congruence rule, ΔADB ≅ ΔADC.

(i) To prove: AD bisects BC
Since ΔADB ≅ ΔADC, by CPCT we get:
BD = CD. Therefore, AD bisects BC. (Hence Proved)

(ii) To prove: AD bisects ∠A
Since ΔADB ≅ ΔADC, by CPCT we get:
∠BAD = ∠CAD. Therefore, AD bisects ∠A. (Hence Proved)

Question 3

Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of ΔPQR. Show that:
(i) ΔABM ≅ ΔPQN
(ii) ΔABC ≅ ΔPQR.

Solution:

Given: AB = PQ, AM = PN (medians), and BC = QR.

Since AM and PN are medians:
BM = ½ BC and QN = ½ QR.
Since BC = QR, we have BM = QN.

(i) To prove: ΔABM ≅ ΔPQN

In ΔABM and ΔPQN:

• AB = PQ (Given)
• AM = PN (Given)
• BM = QN (Proved above)

Therefore, by SSS congruence rule, ΔABM ≅ ΔPQN. (Hence Proved)

(ii) To prove: ΔABC ≅ ΔPQR

Since ΔABM ≅ ΔPQN, by CPCT we get:
∠B = ∠Q.

Now, in ΔABC and ΔPQR:

• AB = PQ (Given)
• ∠B = ∠Q (Proved above)
• BC = QR (Given)

Therefore, by SAS congruence rule, ΔABC ≅ ΔPQR. (Hence Proved)

Question 4

BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.

Solution:

Given: Altitudes BE and CF are equal, so BE = CF and ∠BEC = ∠CFB = 90°.

In right-angled triangles ΔBEC and ΔCFB:

• ∠BEC = ∠CFB = 90° (Given altitudes)
• BC = BC (Common Hypotenuse)
• BE = CF (Given altitude lengths are equal)

By RHS congruence rule, ΔBEC ≅ ΔCFB.

Since the triangles are congruent, their corresponding angles are equal (by CPCT):
∠C = ∠B (or ∠ECB = ∠FBC).

Since the angles opposite to sides AB and AC are equal, the sides themselves must be equal:
AB = AC (Sides opposite to equal angles are equal).

Therefore, ΔABC is an isosceles triangle. (Hence Proved)

Question 5

ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that ∠B = ∠C.

Solution:

Given: An isosceles triangle ABC with AB = AC and AP ⊥ BC (∠APB = ∠APC = 90°).

In right-angled triangles ΔAPB and ΔAPC:

• ∠APB = ∠APC = 90° (By construction)
• AB = AC (Hypotenuse, Given)
• AP = AP (Common Side)

By RHS congruence rule, ΔAPB ≅ ΔAPC.

By CPCT (Corresponding Parts of Congruent Triangles):
∠B = ∠C. (Hence Proved)

💡 Aptitude Corner (Competitive Exams / NTSE Prep)

In competitive examinations like NTSE, Olympiads, and various entrance tests, questions involving altitudes and medians are frequently asked. Here are some key shortcut principles to remember:

• Equal Altitudes Theorem: If two altitudes of a triangle are equal, the triangle is always isosceles. If all three altitudes are equal, the triangle is equilateral.
• RHS application: Always search for the hypotenuse first when applying RHS. If the hypotenuse is a common side or explicitly given as equal, RHS is usually the fastest path to proof.
• Centroid & Median Property: The point where the three medians intersect is called the centroid, and it divides each median in the ratio of 2:1.

Congratulations! More solutions are waiting for you to discover.

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