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: $\displaystyle \widehat{ABD}+\widehat{DAB}=90^{\circ}$
In the triangle ABC: $\displaystyle \widehat{ABD}+\widehat{ACD}=90^{\circ}$
Then $\displaystyle \widehat{DAB}=\widehat{ACD}$
In the same way, $\displaystyle \widehat{ABD}=\widehat{DAC}$
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.
Prove that triangles ABC, ADB, and ADC are similar .Is angle BAD = angle DAC? . noCode:A * *|4 * *3| * * | * * | * * 1 | 2 * B * * * * * * * * * C D
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 $\displaystyle \Delta ABC \sim \Delta ADC$