mathemerize

Solution : Here, \(\triangle\) ABC is an isosceles with AB = AC. \(\therefore\)  \(\angle\) B = \(\angle\) C In \(\triangle\)s ABD and ECF, we have \(\angle\) ABD = \(\angle\) ECF        [\(\because\) \(\angle\) B = \(\angle\) C] \(\angle\) ADB = \(\angle\) EFC = 90 \(\therefore\)  By AA similarity, \(\triangle\) ABD ~ \(\triangle\) ECF

Solution : Given : \(\triangle\) ABC ~ \(\triangle\)  FEG and CD and GH are bisectors of \(\angle\) C and \(\angle\) G respectively. (i) In triangle ACD and FGH  \(\angle\) A = \(\angle\) F         (\(\triangle\) ABC ~ \(\triangle\)  FEG) Since it is given that CD bisects \(\angle\) C and GH bisects \(\angle\) G …

CD and GH are respectively the bisectors of \(\angle\) ACB and \(\angle\) EGF such that D and H lie on sides AB and FE of \(\triangle\) ABC and \(\triangle\) EFG respectively. If \(\triangle\) ABC ~ \(\triangle\) FEG show that (i) \(CD\over GH\) = \(AC\over FG\) (ii) \(\triangle\) DCB ~ \(\triangle\) HGE (iii) \(\triangle\) DCA ~ \(\triangle\) HGF Read More »

Solution : (i)  In triangles, ABC and AMP, we have \(\angle\) ABC = \(\angle\) AMP = 90       (given) \(\angle\)  BAC = \(\angle\) MAP          (common angles) \(\therefore\)  By AA similarity, we have \(\triangle\) ABC ~ \(\triangle\) AMP (ii)  We have : \(\triangle\) ABC ~ \(\triangle\) AMP          …

In the figure, ABC and AMP are two right triangles, right angled at B and M respectively, Prove that (i) \(\triangle\) ABC ~ \(\triangle\) AMP (ii) \(CA\over PA\) = \(BC\over MP\) Read More »

Solution : In triangles ABE and CFB, we have, \(\angle\) AEB = \(\angle\) CBF           (alternate \(\angle\)s are equal) \(\angle\) A = \(\angle\) C                      (opposite angles of parallelogram are equal) \(\therefore\) By AA similarity, \(\triangle\) ABE ~ \(\triangle\) CFB

Solution : (i)  In \(\triangle\) AEP and CDP, we have \(\angle\) AEP = \(\angle\) CDP = 90 \(\angle\) APE = \(\angle\) CPD           (vertically opposite angles) \(\therefore\)  By AA similarity, we have : \(\triangle\) AEP ~ \(\triangle\) CDP (ii)  In \(\triangle\) ABD and CBE, we have \(\angle\) ABD = \(\angle\) CBE    …

In the figure, altitudes AD and CE of \(\triangle\) ABC intersect each other at the point P. Show that : (i) \(\triangle\) AEP ~ \(\triangle\) CDP (ii) \(\triangle\) ABD ~ \(\triangle\) CBE (iii) \(\triangle\) AEP ~ \(\triangle\) ADB (iv) \(\triangle\) PDC ~ \(\triangle\) BEC Read More »

Solution : It is given that \(\triangle\) ABE \(\cong\) \(\triangle\) ACD \(\therefore\)   AB = AC and  AE = AD [ because corresponding parts of congruent triangles are equal ] So,  \(AB\over AD\) = \(AC\over AE\)    or     \(AB\over AC\) = \(AD\over AE\)             ………(1) \(\therefore\)  In triangles ADE and ABC, …

In the figure, if \(\triangle\) ABE \(\cong\) \(\triangle\) ACD, show that \(\triangle\) ADE ~ \(\triangle\) ABC. Read More »

Solution : Given : \(\triangle\) RPQ and \(\triangle\) RTS where \(\angle\) P = \(\angle\) RTS To prove : \(\triangle\) RPQ ~ \(\triangle\) RTS Proof : In \(\triangle\) RPQ and \(\triangle\) RTS Given,    \(\angle\) P = \(\angle\) RTS \(\angle\) R = \(\angle\) R                (common) Hence, By AA similarity, …

S and T are points on sides PR and QR of triangle PQR such that \(\angle\) P = \(\angle\) RTS. Show that \(\triangle\) RPQ ~ \(\triangle\) RTS. Read More »

Solution : Given,  \(QR\over QS\) = \(QT\over PR\) So, \(QT\over QR\) = \(PR\over QS\)               ……….(1) Also given,  \(\angle\) 1 = \(\angle\) 2 Since, sides opposite to equal \(\angle\)s are equal, So , PR = PQ              ………(2) From (1) and (2), we get \(QT\over …

In the figure, \(QR\over QS\) = \(QT\over PR\) and \(\angle\) 1 = \(\angle\) 2. Show that \(\triangle\) PQS ~ \(\triangle\) TQR. Read More »

Solution : Given : A trapezium ABC whose diagonals AC and BD intersect each other at O and AB || DC. To Prove : \(OA\over OC\) = \(OB\over OD\) Proof : In triangle DOC and AOB, AB || DC and AC is transversal, then \(\angle\) DCO = \(\angle\) OAB              …

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 \(OA\over OC\) = \(OB\over OD\). Read More »

Solution : Since BD is a line and OC is a ray on it. \(\angle\) DOC + \(\angle\) BOC = 180 So,  \(\angle\) DOC + 125 = 180 \(\angle\) DOC = 55 In triangle CDO, we have : \(\angle\) CDO + \(\angle\) DOC + \(\angle\) DCO = 180 70 + 55 + \(\angle\) DCO = …

In the figure, \(\triangle\) ODC ~ \(\triangle\) OBA, \(\angle\) BOC = 125 and \(\angle\) CDO = 70. Find the \(\angle\) DOC, \(\angle\) DCO and \(\angle\) OAB. Read More »