In ΔABC, AD⊥BC and AD2=BD×CD. Prove that ∠BAC=90∘.
Answer:
- Given:
AD⊥BC and AD2=BD×CD
Here, we have to find the value of ∠BAC. - Now, we have: AD2=BD×CD⟹BDAD=ADCD
Now, in ΔDBA and ΔDAC, we have
∠BDA=∠ADC=90∘ and BDAD=ADCD∴ ΔDBA∼ΔDAC [By SAS-similarity] As the corresponding angles of similar triangles are equal. So, ∠B=∠2 and ∠1=∠C∴ ∠1+∠2=∠B+∠C⟹ ∠A=∠B+∠C [where, ∠A=∠1+∠2]⟹ 2∠A=∠A+∠B+∠C [Adding ∠A on both sides. ]⟹ 2∠A=∠A+∠B+∠C=180∘ [Sum of angles of a triangle is of 180∘.]⟹ ∠A=∠BAC=90∘.