# right angled triangle

• Oct 15th 2009, 11:07 PM
thereddevils
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 ??
• Oct 16th 2009, 12:59 AM
red_dog
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, $\widehat{ABD}=\widehat{DAC}$
• Oct 16th 2009, 07:56 AM
Soroban
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:

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

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

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