Question

In the adjoining figure ∆ABC & ∆DBC are on the same base BC. AD & BC intersect at O. Prove that area ( ∆ABC)/ area (ΔDBC) = AO /DO .

Answer 1

Two Triangles are said to be similar if their i)corresponding angles are equal and ii)corresponding sides are proportional.(the ratio between the lengths of corresponding sides are equal)

•Similarity of triangles should be expressed symbolically using correct correspondence of their vertices

AA similarity: if two angles of one triangle are respectively equal to two angles of another triangle then the two Triangles are similar.

SOLUTION:

Construction:
Draw AL ⟂ BC and DM ⟂BC .

In ∆ ALO and ∆ DOM
∠ALO = ∠DMO   [each 90°]
∠AOL  = ∠DOM  (vertically opposite angles)
∆ALO ~ ∆ DMO    (By AA Similarity criteria)
AL/DM = AO/DO……………….(1)

[corresponding sides of similar triangles are proportional]
ar(∆ABC)  =1/2 × BC × AL.. ..  ... ...(2)
ar(∆DBC)   = 1/2×BC×DM………….(3)
On dividing eq 2 and 3,
ar(∆ABC)/ar(∆DBC) = 1/2×BC× AL/ ½ × BC ×DM

ar(ABC) / ar (DBC) = AL/DM
ar(ABC) / ar (DBC)  = AO/DO
[from eq 1  ]

HOPE THIS WILL HELP YOU...

Answer 2

Step-by-step explanation:

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