In the diagram, $\angle BAC = \angle ADB = 90{}^\circ$. Prove that $\angle ABD = \angle CAD$, \(

Question

In the diagram, $\angle BAC = \angle ADB = 90{}^\circ$. Prove that $\angle ABD = \angle CAD$, $\Delta ABD$ is similar to $\Delta CAD$, $AD^2 = BD \times CD$.

Accepted Answer

(i) Proved: $\angle ABD=\angle CAD$ (ii) Proved: $\triangle ABD\sim\triangle CAD$ (iii) Proved: $AD^2=BD\times CD$

Proofs in Plane Geometry - Similarity

Proofs in Plane Geometry - Similarity

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