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