Diagonals ac and bd of a trapezium abcd with ab ll dc intersect each other at the point o. using a similarity criterion for two triangles, show that oa\oc =ob\od
Attachments:
Answers
Answered by
2
Answer:
Step-by-step explanation:
below I have attached the answer to the above ques.
Plzz Brainlist me if uh like my solution...
And follow me for More help..
all the best for Boards
.....
Attachments:
manasiriya2003:
Let me know about ur views here
Answered by
0
Answer:
In ΔDOC and ΔBOA,
AB || CD, thus alternate interior angles will be equal,
∴∠CDO = ∠ABO
Similarly,
∠DCO = ∠BAO
Also, for the two triangles ΔDOC and ΔBOA, vertically opposite angles will be equal;
∴∠DOC = ∠BOA
Hence, by AAA similarity criterion,
ΔDOC ~ ΔBOA
Thus, the corresponding sides are proportional.
DO/BO = OC/OA
⇒OA/OC = OB/OD
Hence proved
Similar questions
Environmental Sciences,
6 months ago
Chemistry,
6 months ago
Science,
6 months ago
Computer Science,
1 year ago
English,
1 year ago
Physics,
1 year ago