1. Two triangles are similar if they have: 1) Two shared diagonals of a trapezium intersecting at a point 2) Two angles of one triangle equal to two angles of the other triangle formed by intersecting lines on two sides of an angle 3) Two triangles are similar if their corresponding angles are equal based on another pair of similar triangles

1. CH 6
TRIANGLES
EX 6.3 CONTINUE

2. 3. Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using
a similarity criterion for two triangles, show that ΔOAB ~ ΔOCD
Sol. We have a trapezium ABCD in which AB || DC. The diagonals AC
and BD intersect at O.
In ΔOAB and ΔOCD
AB || DC [Given]
and BD intersects them
∴∠OBA = ∠ODC ...(1) [Alternate angles]
∴Using AA similarity rule,
ΔOAB ~ ΔOCD
similarly,
∠OAB = ∠OCD ...(2)

4. 5. S and T are points on sides PR and QR of ∠P = ∠RTS. Show that ΔRPQ ~ ΔRTS.
Sol. In ΔPQR
T is a point on QR and S is a point on PR such that
∠RTS = ∠P
Now in ΔRPQ and ΔRTS
∠RPQ = ∠RTS
[Given]
∠PRQ = ∠TRS
[Common]
∴ Using AA similarity, we have
ΔRPQ and ΔRTS

5. 6. In the figure, if ΔABE ≌ACD, show that ΔADE ≌ ΔABC.
Sol. We have ΔABE ≌ ΔACD
∴Their corresponding parts are equal,
i.e., AB = AC
AE = AD
∠DAE = ∠BAC [Common]
∴Using SAS similarity, we have
ΔADE ~ ΔABC

6. 7. In the figure, altitudes AD and CE of A ABC intersect each other at the point P. Show that:
(i) ΔAEP ~ ΔCDP
(ii) ΔABD ~ ΔCBE
(iii) ΔAEP ~ ΔADB
(iv) ΔPDC ~ ΔBEC
8. E is a point on the side AD produced of a parallelogram. ABCD and BE intersects CD at F.
Show that ΔABE ~ ΔCFB.
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