Question

in triangle abc , the bisector of ad of a is perpendicular to side bc show that ab = ac and abc is isoceles?

Answer 1

⇰ Given

⇾ In ΔABC

⇾ Bisector AD of A is perpendicular to BC

⇰ To show

⇾ AB = AC

⇾ ΔABC is isosceles triangle

⇰ Solution

⇾ In ΔABC and ΔACD

⇾ ∠BAD = ∠CAD

⇾ AD = AD

⇾ ∠ADB = ∠ADC = 90°

∴ ΔADB ≅ ΔACD (ASA rule) (Angle side angle)

∴ AB = AC (CPCT) (Corresponding parts of congruent triangles)

∵ ΔABC is an isosceles triangle

Answer 2

Answer:

Given : In ∆ ABC, AD is the perpendicular bisector of BC

$\setlength{\unitlength}{20mm}\begin{picture}(16,4)\thicklines\put(8.8,3){\sf\large{A}}\put(7.8,1){\sf\large{B}}\put(10.03,1){\sf\large{C}}\put(8,1){\line(1,0){2}}\put(8,1){\line(1,2){1}}\put(10,1){\line(-1,2){1}}\put(8.9,0.8){\sf\large{D}}\put(8.8,1){\dashbox{0.3}(.2,.2)}\put(9,1){\line(0,1){2}}\end{picture}$

⠀

To Prove : ∆ ABC is an Isosceles Triangle in which AB = AC

⠀

Proof :

$\underline{\bigstar\:\textsf{In \Delta ADB and \Delta ADC :}}$

$:\implies\sf BD=DC\qquad(Divided\:by\:perpendicular)\\\\\\:\implies\sf \angle ADB=\angle ADC\qquad(\angle\:90^{\circ})\\\\\\:\implies\sf AD=AD\qquad(Common)\\\\\\:\implies\sf \Delta \:ADB \cong \Delta \:ADC\qquad(By\: SAS\: theorem)$

⠀

$\sf Now\ as\ \Delta \:ADC \cong \Delta \:ABC\ ,\ hence\\\\\dashrightarrow\sf AB = AC\qquad(By\ CPCT)\\\\\therefore\textsf{ \Delta ABC is an Isosceles Triangle, as Two Sides are Equal}.$