Question

△ABC and △ABD are such that C and D are in opposite half planes and [ABC] = [ABC]. Prove that CD is bisector by AB

Answer 1

Answer:In ΔABC, we have

∠B=2∠C or, ∠B=2y, where ∠C=y

AD is the bisector of ∠BAC. So, let ∠BAD=∠CAD=x

Let BP be the bisector of ∠ABC. Join PD.

In ΔBPC, we have

∠CBP=∠BCP=y⇒BP=PC

In Δ  

′

s ABP and DCP, we have

∠ABP=∠DCP, we have

∠ABP=∠DCP=y

AB=DC     [Given]

and, BP=PC     [As proved above]

So, by SAS congruence criterion, we obtain

ΔABP≅ΔDCP

⇒∠BAP=∠CDP and AP=DP

⇒∠CDP=2x and ∠ADP=DAP=x     [∴∠A=2x]

In ΔABD, we have

∠ADC=∠ABD+∠BAD⇒x+2x=2y+x⇒x=y

In ΔABC, we have

∠A+∠B+∠C=180  

∘

 

⇒2x+2y+y=180  

∘

 

⇒5x=180  

∘

     [∵x=y]

⇒x=36  

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Hence, ∠BAC=2x=72  

∘

 

solution

Step-by-step explanation: