ABC is a triangle in which ∠B=2∠C. D is a point on side BC such that AD bisects ∠BAC and AB =CD. Find m if ∠BAC=m∘
In △BPC, we have
∠CBP=∠BCP=y⟹BP=PC ... (1)
Now, in △ABP and △DCP, we have
∠ABP=∠DCP=y
AB=DC --------------[Given]
and, BP=PC ------[Using (1)]
So, by SAS congruence criterion, we have
△ABP≅△DCP
∴∠BAP=∠CDP=2x and AP=DP,
So in △APD, AP=DP
∠ADP=∠DAP=x
In △ABD, we have
∠ADC=∠ABD+∠BAD⟹3x=2y+x
x=y
In △ABC, we have
∠BAC+∠CBA+∠ACB=180∘
2x+2y+y=180∘
5x=180∘
x=36∘
Hence, ∠BAC=2x=72∘
∴∠BAC=m∘=72∘