solve the 2 question detaily plzzz
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udayraj768oy9yd5:
hmm should think!!1 these are challenging
got the or bit
this is not challenging this is easy
this is really easy want challenging???
ok
what ok want??
yea I want challenging
then check the question which i ask now
then vheck my last question i have 4 good questions
ok
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Hope this Helps
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Given: In ∆ABC and ∆DBE, AC⊥BC, BD ⊥BC and DE ⊥AB. To prove BE/DE = AC/BCProof: In ∆ABC and ∆DBE,‹c = ‹deb 90°angle abc + angle DBE = 90°Angle BDE + angle DBE = angle BD + Angle DBEequal to angle abc equal to angle BDETriangle ABC congrance Triangle DBEequal to AB by BD equal to AC by BE equal to DC by DE are common sideequal to AC by BC equal to BE by DE
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