Question

Que: ΔABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB. Show that ∠BCD is a right angle.​

Answer 1

Solution:

Given, AB = AC and AD = AB

To prove: ∠BCD is a right angle.

Proof:

Consider ΔABC,

AB = AC (Given)

Also, ∠ACB = ∠ABC (Angles opposite to equal sides)

Now, consider ΔACD,

AD = AB

Also, ∠ADC = ∠ACD (Angles opposite to equal sides)

Now,

In ΔABC,

∠CAB + ∠ACB + ∠ABC = 180°

So, ∠CAB + 2∠ACB = 180°

⇒ ∠CAB = 180° – 2∠ACB — (i)

Similarly in ΔADC,

∠CAD = 180° – 2∠ACD — (ii)

Also,

∠CAB + ∠CAD = 180° (BD is a straight line.)

Adding (i) and (ii) we get,

∠CAB + ∠CAD = 180° – 2∠ACB + 180° – 2∠ACD

⇒ 180° = 360° – 2∠ACB – 2∠ACD

⇒ 2(∠ACB + ∠ACD) = 180°

⇒ ∠BCD = 90°

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Hope it will be helpful :)

Answer 2

Answer:

Given:

(i) BE and CF are altitudes.

(ii) AC = AB

To prove:

BE = CF

Proof:

Triangles ΔAEB and ΔAFC are similar by AAS congruency, since;

∠A = ∠A (common arm)

∠AEB = ∠AFC (both are right angles)

AB = AC (Given)

∴ ΔAEB ≅ ΔAFC

and BE = CF (by CPCT).