Question

Is an isosceles aC, Ab=AC and AD perpendicular to BC . Prove that B=C.

Answer 1

$\huge\star\underline\mathfrak\green{Answer:-}$

Step-by-step explanation:

Ans4.In isosceles Triangle ABC

AB= AC [given]

$\angle \: B = \angle \:C \:$

Property of isosceles triangle.

AD = AD [ common]

Thus

$\triangle \: ABD\cong \: \triangle \: ACD$

Thus BD= CD [BY CPCT]

$\angle \: ADB = \angle \:ADC \:$

By CPCT

Now to prove that AD is a perpendicular bisector of BC,we have to prove with the help of linear pair

$\angle \: ADB+ \angle \:ADC =180° \\\\2\angle \: ADB=180°\\\\\angle \: ADB=90°\\\\Thus \angle ADC=90°\\\:$

Hence proved.

Ans 5.

(a) ∆ABC ≈ ∆DBC

AB = DC [Given]

$\angle \: ABC \: = \angle \: BDC\:[Given]$

BC = BC [common]

By AAS CRITERION OF CONGRUENCY OF TRIANGLE ,

$\triangle \: ABC \cong \: \triangle \: DBC$

(ii) If two triangles are congruent than corresponding parts are equal, thus

$\angle \: A = \angle \:D \:$

by CPCT.

(iii) In ∆ AOB and ∆DOC

$AB = DC \: \: \: \: [given]\\ \angle \: A= \angle \: D \: \: [proved \: in \: (ii)] \\ \angle \:AOB = \angle \: DOC \: \: \: \: [vertically \: opposite \: angles]$

Thus by AAS CRITERION OF CONGRUENCY OF TRIANGLES,

$\triangle \: AOB \cong \: \triangle \: DOC$

(iv) Since ∆AOB and ∆ DOC are congruent,so

BO= CO (BY CPCT)

Thus ∆OBC is Isosceles.

Hope it helps you.

Answer 2

$\huge\underbrace\mathfrak \red{ANSWER }$

Ans4.In isosceles Triangle ABC

AB= AC [given]

$\angle \: B = \angle \:C \:$

Property of isosceles triangle.

AD = AD [ common]

Thus

$\triangle \: ABD\cong \: \triangle \: ACD$

Thus BD= CD [BY CPCT]

$\angle \: ADB = \angle \:ADC \:$

By CPCT

Now to prove that AD is a perpendicular bisector of BC,we have to prove with the help of linear pair

$\angle \: ADB+ \angle \:ADC =180° \\\\2\angle \: ADB=180°\\\\\angle \: ADB=90°\\\\Thus \angle ADC=90°\\\:$

Hence proved.

Ans 5.

(a) ∆ABC ≈ ∆DBC

AB = DC [Given]

$\angle \: ABC \: = \angle \: BDC\:[Given]$

BC = BC [common]

By AAS CRITERION OF CONGRUENCY OF TRIANGLE ,

$\triangle \: ABC \cong \: \triangle \: DBC$

(ii) If two triangles are congruent than corresponding parts are equal, thus

$\angle \: A = \angle \:D \:$

by CPCT.

(iii) In ∆ AOB and ∆DOC

$AB = DC \: \: \: \: [given]\\ \angle \: A= \angle \: D \: \: [proved \: in \: (ii)] \\ \angle \:AOB = \angle \: DOC \: \: \: \: [vertically \: opposite \: angles]$

Thus by AAS CRITERION OF CONGRUENCY OF TRIANGLES,

$\triangle \: AOB \cong \: \triangle \: DOC$

(iv) Since ∆AOB and ∆ DOC are congruent,so

BO= CO (BY CPCT)

Thus ∆OBC is Isosceles.

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