How 9 Cheenta students ranked in top 100 in ISI and CMI Entrances?

# Geometry problem - CMI Entrance 2019

## Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.27.4" text_font="Raleway||||||||" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"]let O be a point inside a parallelogram ABCD such that $\angle AOB+\angle COD =180$ prove that $\angle OBC =\angle ODC$

[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.22.4" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px" hover_enabled="0"][et_pb_accordion_item title="Source of the problem" open="off" _builder_version="3.22.4"]

C.M.I (Chennai mathematical institute UG-2019 entrance

[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="3.22.4" open="off"]Geometry

[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.22.4" open="off"]

5 out of 10

[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="4.3.4" hover_enabled="0" open="on"]

challenges and thrills of pre college mathemetics

[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="3.22.7" tab_text_color="#ffffff" body_font="||||||||" tab_font="||||||||" background_color="#ffffff"][et_pb_tab title="Hint 0" _builder_version="3.22.4"]

Do you really need a hint? Try it first!

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.22.7"]

Draw a clear image of the given problem

[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.22.7"]

translate  ABCD along the vector AD SO A' and D are the same , and  so that B' and C are the same

[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.22.7"]

now , $\angle COD +\angle CO'D=\angle COD+\angle A'O'D' =180$

so OCO'D is cyclic . therefore $\angle OO'C =\angle ODC$

[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.22.7"]

Also , vector BC and OO' both equal AD so OBCO' is parallelogram . therefore

$\angle OBC =\angle OO'C=\angle ODC$

## Connected Program at Cheenta

Indian Statistical Institute and Chennai Mathematical Institute offer challenging bachelor’s program for gifted students. These courses are B.Stat and B.Math program in I.S.I., B.Sc. Math in C.M.I.

The entrances to these programs are far more challenging than usual engineering entrances. Cheenta offers an intense, problem-driven program for these two entrances.

## Similar Problem

[/et_pb_text][et_pb_post_slider include_categories="10" _builder_version="3.22.4"][/et_pb_post_slider][et_pb_divider _builder_version="3.22.4" background_color="#0c71c3"][/et_pb_divider][/et_pb_column][/et_pb_row][/et_pb_section]

# Knowledge Partner

### Cheenta. Passion for Mathematics

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.