right angled triangle

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The diagram shows a right angked triangle where angle BAC=90 . A perpendicular is drawn from A to meet the hypothenus BC at D . Prove that triangles ABC , DBA , and DAC are similar .

Is angle BAD = angle DAC ??

No.

In the triangle ADB: $\widehat{ABD}+\widehat{DAB}=90^{\circ}$

In the triangle ABC: $\widehat{ABD}+\widehat{ACD}=90^{\circ}$

Then $\widehat{DAB}=\widehat{ACD}$

In the same way, $\widehat{ABD}=\widehat{DAC}$

Hello, thereddevils!

Quote:

The diagram shows a right-angled triangle where angle BAC=90°.
A perpendicular is drawn from A to meet the hypothenus BC at D.
Prove that triangles ABC, ADB, and ADC are similar .

Code:

A * *|4 * *3| * * | * * | * * 1 | 2 * B * * * * * * * * * C D

Is angle BAD = angle DAC? . no

In right triangle $ABC,\;\angle 1\text{ and }\angle 2$ are complementary: . $\angle 1 + \angle 2 \:=\:90^o$

In right triangle $ADB,\;\angle 1\text{ and }\angle 3$ are complementary: . $\angle 1 + \angle 3 \:=\:90^o$

. . Hence: . $\angle 2 = \angle 3$

So $\Delta ABC$ and $\Delta ADB$ have equal angles:

. . $\begin{array}{ccc}<br /> \Delta ABC & & \Delta ABD \\ \hline<br /> \angle 1 &=& \angle 1 \\<br /> \angle 2 &=& \angle 3 \\<br /> 90^o &=& 90^o \end{array}$

Therefore: . $\boxed{\Delta ABC \:\sim\:\Delta ABD}$

Similarly, prove that $\Delta ABC \sim \Delta ADC$