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In ^@\Delta ABC^@, ^@AD \perp BC^@ and ^@AD^2 = BD \times CD^@. Prove that ^@\angle BAC = 90^\circ^@.
Answer:
- Given:
^@AD \perp BC^@ and ^@AD^2 = BD \times CD^@
Here, we have to find the value of ^@ \angle BAC^@. - Now, we have: ^@AD^2 = BD \times CD \implies \dfrac { BD } { AD } = \dfrac { AD } { CD }^@
Now, in ^@\Delta DBA^@ and ^@\Delta DAC^@, we have
@^ \begin{aligned} &\angle BDA = \angle ADC = 90^\circ \space \text{and} \space \dfrac { BD } { AD } = \dfrac { AD } { CD } \\ \therefore& \space\space\Delta DBA \sim \Delta DAC \space\space\space\space\space\space\space\space\space\space [\text{By SAS-similarity} ] \\ &\text { As the corresponding angles of similar triangles are equal.} \\ &\text{ So, } \space\space \angle B = \angle 2 \text{ and } \angle 1 = \angle C \\ \therefore& \space\space \angle 1 + \angle 2 = \angle B + \angle C \\ \implies&\space\space \angle A = \angle B + \angle C \space\space\space\space\space\space\space\space\space\space [\text{where, } \angle A = \angle 1 + \angle 2 ]\\ \implies&\space\space 2 \angle A = \angle A + \angle B + \angle C \space\space\space\space\space\space\space\space\space\space [\text{Adding}\space \angle A \space \text{on both sides. }]\\ \implies&\space\space 2 \angle A = \angle A + \angle B + \angle C = 180^\circ \space\space\space\space\space\space\space\space\space\space [\text{Sum of angles of a triangle is of}\space 180^\circ.]\\ \implies&\space\space \angle A = \angle BAC = 90^\circ. \end{aligned} @^
