Question

In a right $△$ABC, a perpendicular BD is drawn on to the largest side from the opposite vertex. Which of the following is true with respect to similarity of triangles?

• A. $△$ABD ~ $△$ACB
• B. $△$ABD ~ $△$ABC
• C. $△$ACB ~ $△$ABC
• D. All of these

Answer 1

The correct option is D

All of these

Consider

Consider $△$ABD and $△$ACB:

∠BAD = ∠BAC [common angle]

∠BDA = ∠ABC [ ${90}^{\circ }$]

By AA similarity criterion, $△$ABD ~ $△$ACB ------------(I)

Now Consider $△$ABD and $△$BCD:

∠BAD = ∠BAC = ${90}^{\circ }$ - ∠DCB = ∠DBC [In $△$BCD, ${90}^{\circ }$ - ∠DCB = ∠DBC]

∠BDA = ∠ABC [ ${90}^{\circ }$]

By AA similarity criterion, $△$ABD ~ $△$BCD --------------(II)

From I and II, we have $△$ABD ~ $△$BCD ~ $△$ACB

Hence all the three are correct.

"If a perpendicular is drawn from the vertex of a right angle of a right triangle to the hypotenuse, then the triangles on both sides of the perpendicular are similar to each other and to the complete triangle."