Question

Proof: In AADC, ZADC= 90° given given Seg DB 1 hypt. AC :: BD2 = Х Theorem of geometric mean :: AB u :: AC = AB + BC = U​

Answer 1

Answer:

pleas mark as brainleest

Answer 2

Answer:

, we have

GivenAB=AC and BD=DCTo prove:△≅△ADCProof:In △ADB and △ADC, we haveAB=AC (given)

GivenAB=AC and BD=DCTo prove:△≅△ADCProof:In △ADB and △ADC, we haveAB=AC (given)BD=DC (given)

GivenAB=AC and BD=DCTo prove:△≅△ADCProof:In △ADB and △ADC, we haveAB=AC (given)BD=DC (given)AD=AD (common)

[By SSS congruence property]

2180 o =90 o ∴∠ADB=∠ADC=90 o [from (1)](ii) ∠BAD=∠CAD (∵ corresponding parts of the congruent triangles)

hope it helps

*kimtaehyung21*