Math, asked by rkarjun039, 5 months ago

In the adjoining figure, AB =AC and BD = DC. Prove that ∆ ADB≅ ∆ ADC and hence show that
(i) ∠ADB= ∠ADC=90° (ii)∠BAD= ∠CAD . ​

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rkarjun039: pls help me fast

Answers

Answered by jhariyaaditya0106
17

Answer:

Given

AB=AC and BD=DC

To prove:

△≅△ADC

Proof:

In △ADB and △ADC, we have

AB=AC (given)

BD=DC (given)

AD=AD (common)

∴ △ADB≅△ADC [By SSS congruence property]

(i) ∠ADB=∠ADC (corresponding parts of the congruent triangles ) ...(1)

Now, ∠ADB+∠ADC=180° [∵∠ADB and ∠ADC are on the straight line]

⇒∠ADB+∠ADB=180 ° [from(1)]

⇒2∠ADB=180 °

⇒2∠ADB= 180°/2 =90 °

∴∠ADB=∠ADC=90 °[from (1)]

(ii) ∠BAD=∠CAD (∵ corresponding parts of the congruent triangles)

I hope it help you..........

Answered by roshan200720
4

GIVEN: 1 :- AB=BC

2:- BD=DC

TO PROVE : ∆ADB CONGRUENT TO ∆ADC

PROOF : IN ∆ ADB AND ADC,

AB=AC [GIVEN]

ANGLE ABC = ANGLE ACD [SINCE AB =AC AND SIDE OPPOSITE EQUAL SIDES ARE EQUAL]

BD=DC[GIVEN]

THEREFORE, BY SAS CONGRUENCEY

∆ADB CONGRUENT TO ∆ACD

THEREFORE, ANGLE ADB = ANGLE ACD[C.P.C.T]

ANGLE BAD = ANGLE CAD[C.P.C.T]

NOW,

ANGLE ADB + ANGLE ADC =180

∠ADB + ∠ ADB = 180

2∠ADB =180

∠ADB=90

THEREFORE, ∠ADB= ∠ADC=90°


sanjaysanskar411: good answer
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