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 ΔDACandΔDEC,∠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∘InΔBED,∠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