Question

If a perpendicular is drawn from the vertex containing the right angle of a right triangle to the hypotenuse then prove that the triangle on each side of the perpendicular are similar is product of the lengths of the two parts of the hypotenuse.

Answer 1

$△ABC$ right angled at $B$ and $\perp$ from $B$ intersecting $AC$ at $D\left(BD\perp AC\right)$

$In△ADB\&△ABC$

$\angle A=\angle A$(common)

$\angle ADB=\angle ABC${ Each ${90}^{\circ }$}

$△ADB\sim △ABC${ By AA Similarity criterion} -(1)

Similarly,

$In△BDC\&△ABC$

$\angle C=\angle C${ common}

$\angle BDC=\angle ABC${ Each ${90}^{\circ }$}

$△BDC\sim △ABC${ By AA Similarity Criterion} -(2)

from$\left(1\right)\&\left(2\right)$

$△ADB\sim △ABC\&△BDC\sim △ABC$

If one $△le$ is similar to another $△le$,and second$△le$ is similar, to the third $△le$,then first & third $△le$ are similar.

$△ADB\sim △BDC$