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Question

# Question 18 If ABC is a right angled triangle such that AB = AC and bisector of angle C intersects the side AB at D, then prove that AC + AD = BC.

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Solution

## Given: In right angled ΔABC,AB=AC and CD is the bisector of ∠C. Construction: Draw DE ⊥ BC. To prove: AC + AD = BC Proof: In right angled ΔABC,AB=AC and BC is a hypotenuse [given] ∠A=90∘ In ΔDAC and ΔDEC, ∠A=∠3=90∘ ∠1=∠2 [given, CD is the bisector of ∠C] DC = DC [common sides] ∴ΔDAC≅ΔDEC [by AAS congruence rule] ⇒DA=DE. [by CPCT] ...(i) and AC = EC ...(ii) In ΔABC,AB=AC ∠C=∠B [angles opposite to equal sides are equal] ...(iii) Again, in ΔABC,∠A+∠B+∠C=180∘ [by angle sum property of a triangle] ⇒90∘+∠B+∠B=180∘ [from Eq. (iii)] ⇒2∠B=180∘−90∘⇒2∠B=90∘⇒∠B=45∘In ΔBED, ∠5=180∘−(∠B+∠4) [by angle sum property of a triangle] =180∘−(45∘+90∘)=180∘−135∘=45∘∴∠B=∠5 ⇒DE=BE[∵ sides opposite to equal angles are equal]...(iv) From Eqs. (i) and (iv). DA = DE = BE ...(v) ∵BC=CE+EB = CA + DA [from Eqs. (ii) and (v)] ∴AD+AC=BC

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