In the figure, ΔABC and ΔABD are two triangles on the same base AB. If line segment CD is bisected by `bar(AB)` at O, show that ar `(DeltaABC) = ar (DeltaABD).`

Triangle ABC and parallelogram ABEF are on the same base, AB as in between the same parallels AB and EF. Prove that ar (DeltaABC) = 1/2 ar(|| gm ABEF)

In the figure, ABC is a triangle in which AB= AC. Points D and E are points on the side AB and AC respectively such that AD=AE. Show that points B, C, E and D lie on a same circle.

In the given figure two sides AB and BC and median AM of one triangle ABC are respectively equal to side PQ and QR and median PN of DeltaPQR . Show the DeltaABC~=DeltaPQR .

In the given Figure AB and CD are intersecting at ‘O’, OA = OB and OD = OC. Show that (i) Delta AOD ~= DeltaBOC and (ii) AD || BC .

A line cuts two sides AB and side AC of DeltaABC in point P and Q respectively. Show that (A(DeltaABC))/((A(DeltaAPQ)))=(APxxAQ)/(ABxxAC)

Procedure : (i) Draw a circle with centre O and with suitable radius. (ii) Make it a semi - circle through folding . Consider the poin A,B on it (iii) Make crease along AB in the semi circles and open it. (iv) We get one more crease line on the another part of semi circle, name it as CD (observe AB= CD) (v) Join the radius to get the DeltaOAB and DeltaOCD . (vi) Using trace paper, take the replicas of triangle DeltaOAB and DeltaOCD (vii) Place these triangles DeltaOAB and DeltaOCD one on the other. Observation : (i) What do you observe ? Is DeltaOAB-=DeltaOCD ? (ii) Construct perpendicular line to the chords AB and CD passing through the centre O. Measure the distance from O to the chords

In the adjoining figure, O is the centre and seg AB is a diameter. At point C on the circle, the tangent CD is drawn. Line BD is tangent at B . Prove that seg OD || seg AC.

Let A(p^2,-p),B(q^2, q),C(r^2,-r) be the vertices of triangle ABC. A parallelogram AFDE is drawn with D,E, and F on the line segments BC, CA and AB, respectively. Using calculus, show that the maximum area of such a parallelogram is 1/2(p+q)(q+r)(p-r)dot