1. ## right angled triangle

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 ??

2. No.

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

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

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

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

3. Hello, thereddevils!

The diagram shows a right-angled triangle where angle BAC=90°.
A perpendicular is drawn from A to meet the hypothenus BC at D.
Code:
                A
*
*|4 *
*3|     *
*  |        *
*   |           *
* 1  |            2 *
B *  *  *  *  *  *  *  *  * C
D
Is angle BAD = angle DAC? . no

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

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

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

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

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

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

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