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Standard X

Mathematics

Perpendicular from Right Angle to Hypotenuse Divides the Triangle into Two Similar Triangles

In ΔABC, AD...

Question

In $Δ$ ABC, AD is the perpendicular bisector of BC(see adjacent figure). Show that $Δ$ ABC is an isosceles triangle in which AB $=$ AC.

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Solution

$\begin{matrix} A D ⊥_{r} b i sec t o r o f B C \Rightarrow∴ B D = D C \Rightarrow∴ ∠ B A D = ∠ C A B = θ H e r e ∠ A B C = ∠ C A B = (90^{o} - θ) T h e n, A B = A C [T w o a n g l e s a r e e q u a l s] \end{matrix}$

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Question 2
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In ΔABC, AD is the perpendicular bisector of BC (see the given figure). Show that ΔABC is an isosceles triangle in which AB = AC.

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In ΔABC, AD is the perpendicular bisector of BC(see adjacent figure). Show that ΔABC is an isosceles triangle in which AB=AC.

AD⊥rbisectorofBC⇒∴BD=DC⇒∴∠BAD=∠CAB=θHere∠ABC=∠CAB=(90o−θ)Then,AB=AC[Twoanglesareequals]

In $Δ$ ABC, AD is the perpendicular bisector of BC(see adjacent figure). Show that $Δ$ ABC is an isosceles triangle in which AB $=$ AC.

$\begin{matrix} A D ⊥_{r} b i sec t o r o f B C \Rightarrow∴ B D = D C \Rightarrow∴ ∠ B A D = ∠ C A B = θ H e r e ∠ A B C = ∠ C A B = (90^{o} - θ) T h e n, A B = A C [T w o a n g l e s a r e e q u a l s] \end{matrix}$