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Triangles

Short Answer Type

191. 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

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192.

In the given Fig. $<pre>uncaught exception: mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56 #0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>$ and ∠1 = ∠2, show that ∆PQS ~ ∆TQR.

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193. S and T are points on sides PR and QR of ∆PQR such that ∠P = ∠RTS. Show that ∆RPQ ~ ∆RTS.

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194.

In the given fig, if $increment ABE approximately equal to increment ACD comma$ show that $increment ADE tilde increment ABC$ .

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Long Answer Type

195.

In the given Fig, altitudes AD and CE of $increment ABC$ intersects each other at the point P. Show that:
(i) ∆AEP ~ ∆CDP
(ii) ∆ABD ~ ∆CBE
(iii) ∆AEP ~ ∆ADB
(iv) ∆PDC ~ ∆BEC.

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Short Answer Type

196. E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ∆ABE ~ ∆CFB.

Given : E is a point on side CD produced of parallelogram ABCD. BE intersects CD at F.
To prove: ∆ABE ~ ∆CFB.

Proof: In ∆ABE and ∆CFB, we have,
∠AEB = ∠CBF [Alternate angles]
and ∠A = ∠C
[Opposite angles of a parallelogram]
Therefore, by using AA similar condition,
∆ABE ~ ∆CFB.

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197.

In the given fig, ABC and AMP are two right triangles, right angled at B and M respectively.
Prove that:
(i) $increment ABC tilde increment AMP$
(ii) $space CA over PA equals BC over MP.$

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Long Answer Type

198.

CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ΔABC and ΔEFG respectively. If ΔABC ~ ΔFEG, show that:
(i) $CD over GH equals AC over FG$
(ii) $increment DCB tilde increment HGE$
(iii) $space space increment DCA tilde increment HGF.$